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Results & Lemmas (22)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Lemma 1 Lemma 1 ([14]) Let an be a sequence of nonnegative real numbers such that a1 = 1, and that for n ≥2 the sequence an is a convex decreasing,…
Lemma 1 ([14]) Let {an} be a sequence of nonnegative real numbers such that a1 = 1, and that for n ≥2 the sequence {an} is a convex decreasing, i.e., 0 ≥an+2 – an+1 ≥an+1 – an, for all n ∈N. Then, Re ∞  n=1 anzn–1
Lemma 2 Lemma 2 ([14]) If An ≥0, nAn and nAn – (n + 1)An+1 are both nonincreasing, then the function f (z) = z + ∞ n=2 Anzn is in S∗.
Lemma 2 ([14]) If An ≥0, {nAn} and {nAn – (n + 1)An+1} are both nonincreasing, then the function f (z) = z + ∞ n=2 Anzn is in S∗.
Lemma 3 Lemma 3 ([24]) Let f (z) = z + ∞ n=2 Anzn. Suppose that 1 ≥2A2 ≥··· ≥(n + 1)An+1 ≥··· ≥0 or 1 ≤2A2 ≤··· ≤(n + 1)An+1 ≤··· ≤2, then f is…
Lemma 3 ([24]) Let f (z) = z + ∞ n=2 Anzn. Suppose that 1 ≥2A2 ≥··· ≥(n + 1)An+1 ≥··· ≥0 or 1 ≤2A2 ≤··· ≤(n + 1)An+1 ≤··· ≤2, then f is close-to-convex with respect to the convex function –log(1 – z) in D.
Lemma 4 Lemma 4 ([24]) Suppose that f is an odd function (i.e., of the form f (z) = z + ∞ n=2 A2n–1z2n–1), such that 1 ≥3A3 ≥··· ≥(2n + 1)A2n+1…
Lemma 4 ([24]) Suppose that f is an odd function (i.e., of the form f (z) = z + ∞ n=2 A2n–1z2n–1), such that 1 ≥3A3 ≥··· ≥(2n + 1)A2n+1 ≥··· ≥0 or 1 ≤3A3 ≤··· ≤(2n + 1)A2n+1 ≤··· ≤2, then f is close-to-convex with respect to the convex function (1/2)log((1 + z)/(1 – z)).
Lemma 5 Lemma 5 ([1]) Let An ∞ n=1 be a sequence of nonnegative real numbers such that A1 = 1 and (n + 1)An+1 ≤nAn;(2n)2A2n ≤(2n – 1)2A2n–1 for all…
Lemma 5 ([1]) Let {An}∞ n=1 be a sequence of nonnegative real numbers such that A1 = 1 and (n + 1)An+1 ≤nAn;(2n)2A2n ≤(2n – 1)2A2n–1 for all n ∈N. Then, the functions defined by the series n k=1 Akzk and ∞ n=1 Anzn are convex in the direc- tion of the imaginary axis (see for details [1, Theorem 2.3.5, p. 34]).
Lemma 6 Lemma 6 ([21]) Let  ⊂C, and suppose that ψ: C3 × D →C satisfies the condition ψ(is,t,u + iv;z) /∈ for z ∈D and for real s,t,u and v…
Lemma 6 ([21]) Let  ⊂C, and suppose that ψ : C3 × D →C satisfies the condition ψ(is,t,u + iv;z) /∈ for z ∈D and for real s,t,u and v satisfying t ≤–  1 + s2 /2 and t + u ≤0. If p(z) is analytic in D, with p(0) = 1 and ψ(p(z),zp′(z),z2p′′(z);z) ∈ for z ∈D, then Re(p(z)) > 0, for all z ∈D. This lemma is a special case of Theorem 1 due to Miller and Mocanu in [21].
Lemma 7 Lemma 7 ([34]) For α < 1, β < 1, we have P(α) ∗P(β) ⊂P(δ), where δ = 1 – 2(1 – α)(1 – β). The value of δ is the best possible. In this…
Lemma 7 ([34]) For α < 1, β < 1, we have P(α) ∗P(β) ⊂P(δ), where δ = 1 – 2(1 – α)(1 – β). The value of δ is the best possible. In this paper, we study certain geometric properties such as the close-to-convexity, star- likeness, and convexity of the function z1F2(a;b,c;z). We also study the boundedness prop- erty of the function z1F2(a;b,c;z) in the concluding section. Several special cases and corollaries of our main results are also pointed out. 3 Close-to-convexity of z1F2(a;b,c;z) This sectio
Theorem 1 Theorem 1 Let a,b,c > 0 and a ≤bc/2, then the function z1F2(a;b,c;z) is close-to-convex with respect to the convex function –log(1 – z).
Theorem 1 Let a,b,c > 0 and a ≤bc/2, then the function z1F2(a;b,c;z) is close-to-convex with respect to the convex function –log(1 – z).
Corollary 1 Corollary 1 Let a,b,c > –1, a + 1 ≤(b + 1)(c + 1)/2 and a ̸= 0, then z1F′ 2(a;b,c;z) is univa- lent in D.
Corollary 1 Let a,b,c > –1, a + 1 ≤(b + 1)(c + 1)/2 and a ̸= 0, then z1F′ 2(a;b,c;z) is univa- lent in D.
Corollary 2 Corollary 2 Let c ≥1, then z0F′ 1(–;c;z) is univalent in D.
Corollary 2 Let c ≥1, then z0F′ 1(–;c;z) is univalent in D.
Theorem 2 Theorem 2 Let a,b,c > 0 and a ≤4bc/3, then z1F2(a;b,c;z2/4) is close-to-convex with re- spect to (1/2)log((1 + z)/(1 – z)).
Theorem 2 Let a,b,c > 0 and a ≤4bc/3, then z1F2(a;b,c;z2/4) is close-to-convex with re- spect to (1/2)log((1 + z)/(1 – z)).
Theorem 3 Theorem 3 Let b,c > 0, and 3bc 5bc + 8b + 8c + 8 ≤a ≤bc/4. Then, the function z1F2(a;b,c;z) is in the class KS∗.
Theorem 3 Let b,c > 0, and 3bc 5bc + 8b + 8c + 8 ≤a ≤bc/4. Then, the function z1F2(a;b,c;z) is in the class KS∗.
Corollary 3 Corollary 3 Let c ≥4,then the function z0F1(–;c;z) is in the class KS∗. 5 Convexity of z1F2(a;b,c;z)
Corollary 3 Let c ≥4,then the function z0F1(–;c;z) is in the class KS∗. 5 Convexity of z1F2(a;b,c;z)
Theorem 4 Theorem 4 Let b,c > –1,abc ̸= 0 and –26 + 10b + 10c + 2bc 29 + 13b + 13c + 5bc ≤a ≤bc + b + c – 3 4, then 1F2(a;b,c;z) is a convex function…
Theorem 4 Let b,c > –1,abc ̸= 0 and –26 + 10b + 10c + 2bc 29 + 13b + 13c + 5bc ≤a ≤bc + b + c – 3 4 , then 1F2(a;b,c;z) is a convex function in D.
Theorem 5 Theorem 5 Let a,b,c > 0 and a ≤bc/4, then z1F2(a;b,c;z) is convex in the direction of the imaginary axis.
Theorem 5 Let a,b,c > 0 and a ≤bc/4, then z1F2(a;b,c;z) is convex in the direction of the imaginary axis.
Theorem 6 Theorem 6 Let c be a real number such that c ≥3 – 3β + 2β2 2(1 – β), (11) and 0F′ 1(–;c;z) ̸= 0, then 0F1(–;c;z) is a convex function of…
Theorem 6 Let c be a real number such that c ≥3 – 3β + 2β2 2(1 – β) , (11) and 0F′ 1(–;c;z) ̸= 0, then 0F1(–;c;z) is a convex function of order β (0 ≤β < 1).
Corollary 4 Corollary 4 Let c be a real number such that c ≥5 – 5β + 2β2 2(1 – β), and 0F′ 1(–;c – 1;z) ̸= 0, then z0F1(–;c;z) ∈S∗(β) for 0 ≤β < 1. If…
Corollary 4 Let c be a real number such that c ≥5 – 5β + 2β2 2(1 – β) , and 0F′ 1(–;c – 1;z) ̸= 0, then z0F1(–;c;z) ∈S∗(β) for 0 ≤β < 1. If we let f (z) = z0F1(–;c;z) and h(z) = f (z2/4)/(z/4), then we have zh′(z) h(z) = 2z2 4 f ′(z2/4) f (z2/4) – 1. (13) This observation and Corollary 4 immediately yields the following result.
Corollary 5 Corollary 5 Let c be a real number such that c ≥5 – 5β + 2β2 2(1 – β), 1/2 ≤β < 1, and 0F′ 1(–;c – 1;z) ̸= 0, then z0F1(–;c;z2/4) ∈S∗(2β –…
Corollary 5 Let c be a real number such that c ≥5 – 5β + 2β2 2(1 – β) , 1/2 ≤β < 1, and 0F′ 1(–;c – 1;z) ̸= 0, then z0F1(–;c;z2/4) ∈S∗(2β – 1).
Theorem 7 Theorem 7 Let a,b,c > 0, a ≤bc, and 2bc(b + 1)(c + 1) ≥a  4(b + 1)(c + 1) – (a + 1) , (14) then Re 1F2(a;b,c;z) > 1/2 for z ∈D.
Theorem 7 Let a,b,c > 0, a ≤bc, and 2bc(b + 1)(c + 1) ≥a  4(b + 1)(c + 1) – (a + 1)  , (14) then Re{1F2(a;b,c;z)} > 1/2 for z ∈D.
Corollary 6 Corollary 6 Let c ≥(1 + √ 7)/2, then Re 0F1(–;c;z) > 1/2 for z ∈D.
Corollary 6 Let c ≥(1 + √ 7)/2, then Re{0F1(–;c;z)} > 1/2 for z ∈D.
Theorem 8 Theorem 8 Let a,b,c > 0, a ≤bc, c > a – b – 1 and 2bc(b + 1)(c + 1) ≥a  4(b + 1)(c + 1) – (a + 1) . If f ∈R, then the convolution…
Theorem 8 Let a,b,c > 0, a ≤bc, c > a – b – 1 and 2bc(b + 1)(c + 1) ≥a  4(b + 1)(c + 1) – (a + 1)  . If f ∈R, then the convolution z1F2(a;b,c;z) ∗f (z) is in H∞∩R.
Corollary 7 Corollary 7 Let c ≥1+ √ 7 2 and f ∈R, then the convolution z0F1(–;c;z) ∗f (z) is in H∞∩R. Acknowledgements The fourth author was supported…
Corollary 7 Let c ≥1+ √ 7 2 and f ∈R, then the convolution z0F1(–;c;z) ∗f (z) is in H∞∩R. Acknowledgements The fourth author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2019R1I1A3A01050861). Funding Not applicable. Availability of data and materials Not applicable. Declarations Competing interests The authors declare no competing interests.
Function classes studied:

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